Chapter 23 Linear Motion

Chapter 23

The simplest example of a parametrized curve arises when studying the motion of an object along a straight line in the plane. We will start by studying this kind of motion when the starting and ending locations are known.

23.1 Motion of a Bug Example 23.1.1. A bug is spotted at ¥§¦©¨ in the -

P=(2,5) ¦¨ plane. The bug walks in a straight line fromPSfrag¥ toreplacements at a constant speed . It takes the bug 5 seconds to reach

-axis ¡

. Assume the units of our coordinate system are feet.

Q=(6,3) ¥ What is the speed of the bug along the line connecting

-axis and ? Compute the horizontal and vertical speeds of the -axis bug and show they are both constant.

Solution. A standard technique in motion problems is to (a) A bug walking from ¢

to £ . analyze the and -motion separately. This means we

look at the projection of the bug location onto the and

-axis separately, studying how each projection moves. light on bug ¤ -axis casts shadow We can think of these projections as “shadows” cast by a for x−motion R=(x ,y ) light on bug ﬂashlight onto the two axes: PSfrag replacements shadow 1 1 on y−axis casts shadow

P=(2,5) for y−motion For the -motion, we study the “shadow” on the -axis 5

S=(x ,y ) which starts at “2” and moves toward “6” on the -axis. 3 2 2

Q=(6,3)

For the -motion, we study the “shadow” on the -axis -axis 2 6 -axis shadow on x−axis which starts at “5” and moves toward “3” along the -axis. In general, speed is computed by dividing distance by time elapsed, so (b) How to view motion in

# the coordinates. ¨ %$& ')(*¨ %$+,'

dist ¨ ¥ !" ft ft

¦ ¦ sec sec (23.1) Figure 23.1: Visualizing the

model for the bug problem.

,- feet

¦ .

sec This is the speed of the bug along the line connecting ¥ and . 315 316 CHAPTER 23. LINEAR MOTION

The hard part of this problem is to show that the speed of the horizontal and vertical shadows are also constant. This might seem obvious

when you ﬁrst think about it. In order to actually show it, let’s take two

¦©¨ ¢¡ £¡ ¤&¦ ¨

intermediary positions and along the bugs path.

' '

¦¡

We are going to relate the horizontal speed ¥ of the bug between and

¥ §¡

, the vertical speed of the bug between and and the speed of

' ' ¤

the bug from to . Actually, because there are positive or negative di-

rections for the and axes, we will allow horizontal and vertical “speed” ©

to be a ¨ quantity, with the obvious meaning. If it takes seconds for

¤ ©

the bug to travel from to , then is the elapsed time for the horizontal

¢¡

motion from to and also the elapsed time for the vertical motion from

£¡ ¥ ¦ ¨ $ ¢¡

to . The horizontal speed is the directed distance di-

' '

¤ ¥

vided by the time elapsed © , whereas the vertical speed is the directed

¦ ¨ $ £¡ ©

distance divided by the time elapsed .

¤ ¥ We want to show that ¥ and are both constants! To do this, we

have these three equations:

$ ¢¡

¥ ¦ ¦

$ £¡

¥ ¦ ¦

¨ $ ¡ ' ( ¨ $ £¡ '

¨ ) ¤ ( '

distance '

' '

¦ ¦ ¦

Now, square each side of the three equations and combine them to con-

' ' ' ' ¤ ' ' '

us the key equation:

' ' '

¦¥ (¥

(23.2) .

On the other hand, the ratio of the vertical and horizontal speed gives

¦ ¦

$ £¡

¥ ¦

¦ “slope of line connecting and ”

$ ¡

' (23.3)

¦ $

¤ ¥

Solving for ¥ in terms of we can write

¦ $ ¥

¥ (23.4)

Since ¥ is positive,

- -

#" '&

¦ ¥ ' ( ¥ ' ¦ ¥ ' (! $ ¥ ¦ ¥ ¦ ¥

$ %$ 23.2. GENERAL SETUP 317

that is,

feet

¥ ¦ ¦ ¦¤£ ¦¦¥

¢ .¨§ .

- ¡

- sec

¡ ft

¦ $ ¥ ¦ $

By (23.4), ¥ sec . This shows the speed in both the hori-

. ¤

zontal and vertical directions is constant as the bug moves from to .

¤ ¥ Since and were any two intermediary points between and , the horizontal and vertical bug speeds are constant.

Example 23.1.2. Parametrize the motion in the previous problem.

¦¦¥ ¥ ¦ $ ¥

Solution. We have shown ¥ feet/sec and feet/sec, so

.¨§ .

¨ ¦ ¨ -coordinate of the bug at time

distance traveled in

¦ ¨ (

beginning -coordinate the -direction in ¢

seconds

¦ (¥

.§ .

¨ ¦ ¨ -coordinate of the bug at time

distance traveled in

( ¦ ¨

beginning -coordinate the -direction in ¢

seconds

¦ $¥

. .

¨¥, ¦¨ ( ¨¥ ¥ %$*¨¥ ¥,)¦¨

As a check, notice that ¥ is the starting

.§ .

¨ ¦ ¨ ( ¨¥ $ ¨¥ ¦ ¨

location and ¥ is the ending location.

.§ .

23.2 General Setup

¥ ¦ ¨ ¦¡ £¡ ¦ ¨

Given two points and in the plane, ' ' -axis ending we can study motion of an object along the line connecting PSfrag replacements location

Q=(x ,y ) ¥ and . In so doing, you need to ﬁrst specify the starting 2 2 starting location

location and the ending location of the object; lets say we -axis start at ¥ and proceed to . Fix the distance units used -axis moving from P to Q at in the coordinate system (feet, inches, miles, meters, etc.) P=(x1 ,y1 ) a constant speed and the time units used (seconds, hours, years, etc.). As Figure 23.2: Typical sce-

highlighted in the solution of Example 23.1.2, the key is nario. to analyze the -motion and -motion separately.

We will be imposing an assumption that the speed along the line connecting ¥ and in Figure 23.1(b) is a constant. There is a crucial ob- servation we need to make, a special case of which was the content of Example 23.1.1.

Important Facts 23.2.1 (Linear motion). Assume that an object moves

from ¥ to along a straight line at a constant speed , as in Figure 23.2.

318 CHAPTER 23. LINEAR MOTION

¥ Then the speed ¥ in the -direction and the speed in the -direction are

both constant. We also have two useful formulas:

' ' '

¦ ¥ ( ¥

¤ slope of line of travel,

¥ ¡

¦ when the line is non-

¥ vertical. This fact is established using the same reasoning as in Example 23.1.1.

Let’s make a few comments. To begin with, if the line of travel is either

¦ ¥ ¥ ¦ ¥ PSfrag replacements vertical or horizontal, then either ¥ or and Fact 23.2.1 isn’t

!!! really saying anything of interest. -axis CAUTION -axis !!! The formulas in Fact 23.2.1 only work for linear motion.

-axis ¤ ¤ -axis -axis Q

P Q

PSfrag replacements no y−motion -axis

-axis no x−motion -axis P

Figure 23.3: Horizontal or vertical motion.

For any other line of travel, we can use the reasoning used in Exam-

ple 23.1.1. Pay attention that the horizontal speed ¥ and the vertical ¤

speed ¥ are both directed quantities; i.e. these can be positive or nega-

tive. The sign of ¥ will indicate the direction of motion: If is positive,

then the horizontal motion is to the right and if ¥ is negative, then the ¤ horizontal motion is to the left. Similarly, the sign of ¥ tells us if the vertical motion is upward or downward.

Returning to Figure 23.2, to describe the -motion, two pieces of in-

formation are needed: the starting location (in the -direction) and the

constant speed ¥ in the -direction. So,

¨ ¦ ¨ -coordinate of the object at time

distance traveled in

¦ ¨ (

beginning -coordinate the -direction in ¢

time units

£¢

¦§¢¡ ( ¥ . 23.2. GENERAL SETUP 319

If we are not given the horizontal velocity directly, rather the time ©

required to travel from ¥ to , then we could compute using the fact

¢¡ "¦

that the object starts at "¦ and travels to :

' ¨

directed horizontal distance traveled

¥ ¦

¨ time required to travel this distance

$ ¨

¨ ending -coordinate starting -coordinate

¨ time required to travel this distance

$ ¡

¥ ¦ ¦

To describe the -motion in Figure 23.2, we proceed similarly. We will

denote by ¥ the constant vertical speed of the object, then after time

¤ ¢

units the object has traveled ¥ units. So,

¨ ¦ ¨ -coordinate of object at time

distance traveled in

¦ ¨ (

beginning -coordinate the -direction in ¢

time units

¤ ¢

¦ £¡ ( ¥ . In summary,

Important Fact 23.2.2 (Linear motion). Suppose an object begins at a

¦ ¨ ¡ £¡

point ¥ and moves at a constant speed along a line connecting

¦ ¨

¥ to another point . Then the motion of this object will trace out

' '

a line segment which is parametrized by the equations:

£¢

¦ ¨ ¦ ¡ (¥

¤ ¢

¦ ¨ ¦+£¡ ( ¥

. ¤ Return to the linear motion problem stud- -axis location E where Example 23.2.3. PSfrag replacements bug enters region

ied in Example 23.1.1 and 23.1.2. However, now assume P=(2,5) ¨ that the point *¦ is located at the center of a circular circular region region of radius 1 ft. When and where does the bug enter radius 1 ft. this circular region? -axis Q=(6,3) -axis

Solution. The parametric equations for the linear motion Figure 23.4: A bug crosses a

of the bug are given by: circular boundary.

"¦ ¨ ¦ ( ¥

.§

¦+ ¨ ¦ $ ¥

. .

The equation of the boundary of the circular region centered at is given

' '

¨ $ , ( ¨ $& ¦

. 320 CHAPTER 23. LINEAR MOTION

To ﬁnd where and when the bug crosses into the circular region, we

determine where and when the linear path and the circle equation have

a simultaneous solution. To ﬁnd such a location, we simply plug "¦

¦ ¨

and into the circle equation:

' '

¨ ¨ $ , ( ¨ ¨ $ ¦

' '

$ ¦ ¨ %( ¥ $ , ( ¨ $¥

.¨§ .

' '

(¥ ( $ ¦ $ (

£ £ £

. . . .

¥ $ ( ¦ ¥

.¨§ § .

Notice, this is an equation in the single variable . Finding the solution of this equation will tell us when the bug crosses into the region. Once we know when the bug crosses into the region, we can determine the location by plugging this time value into our parametric equations. By

the quadratic formula, we ﬁnd the solutions are

¨ ¨¥ $ ¨

£ £

§ .§

¦ ¦

. .§ §,.

We know that the bug reaches the point in 5 seconds, so the sec-

ond solution seconds is the time when the bug crosses into

§ § the circular region.. (If the bug had continued walking in a straight

line directly through , then the time when the bug leaves the circu-

lar region would correspond to the other solution seconds.)

Finally, the location ¡ of the bug when it crosses into the region is

¦ ¨ ¨ ¨ ¦ ¨ ¥

¡ .

.¨§ § .¨§ § . .